Proving a the square root of a function to be Riemann integrable How could I prove that the square root of a Riemann integrable function $f$ on a given interval, where $f(x) > 0$ is also Riemann integrable?
 A: You can use the following result:
Suppose $f$ is Riemann integrable in $\,[a,b]\,$ and $\,f\big([a,b]\big)\subset \left[m,M\right]$. If $g$ is continuous in $\,\left[m,M\right]\,$ then $\,h=g\circ f\,$ is Riemann integrable in $\,\left[a,b\right]$.
A: hint: $|\sqrt{f(u)}-\sqrt{f(v)}|=\dfrac{|f(u)-f(v)|}{\sqrt{f(u)}+\sqrt{f(v)}}\leq \dfrac{1}{\sqrt{m}}\cdot |f(u)-f(v)|$, with $m = \inf\{f(x): x \in [a,b]\}$
A: The most straightforward way is Reveillark's, the validity of which follows immediately from Lebesgue's criterion for Riemann-integrability.
Here is a direct proof that doesn't appeal to this result. 
Let $f \geq 0$ be Riemann-integrable on $[a,b]$. Let $\varepsilon > 0$ be given. Since $f$ is integrable, there exist step functions $\varphi$, $\psi$ such that
$$0 \leq \varphi \leq f \leq \psi, \qquad \int_a^b (\psi - \varphi) \leq \varepsilon^2/(b-a).$$
Now $\sqrt{\varphi}, \sqrt{\psi}$ are step functions satisfying $\sqrt{\varphi} \leq \sqrt{f} \leq \sqrt{\psi}$ and
$$\int_a^b \left( \sqrt{\psi} - \sqrt{\varphi} \right) \leq \int_a^b \sqrt{\psi - \varphi} \leq (b-a)^{1/2} \left[\int_a^b (\psi - \phi)\right]^{1/2} \leq \varepsilon,$$
where the first inequality follows from $\sqrt{\psi} - \sqrt{\varphi} \leq \sqrt{\psi - \varphi}$ and the second follows from Cauchy-Schwarz. (Note that the argument is not circular because $\sqrt{\psi - \varphi}$ is a step function.)
Since $\varepsilon > 0$ was arbitrary, this shows that $\sqrt{f}$ is integrable.
